Answer:
The range of [H⁺] is from 2.51 x 10⁻⁶ M to 6.31 x 10⁻⁶ M,
Explanation:
To answer this problem we need to keep in mind the <u>definition of pH</u>:
So now we <u>calculate [H⁺] using a pH value of 5.2 and of 5.6</u>:
-5.2 = log [H⁺]
= [H⁺]
6.31 x 10⁻⁶ M = [H⁺]
-5.6 = log [H⁺]
= [H⁺]
2.51 x 10⁻⁶ M = [H⁺]
Answer:
Fe(CN)₂, FeCO₃, Pb(CN)₄, Pb(CO₃)₂
Explanation:
Cations (positively charged ions) can only form ionic bonds with anions (negatively charged ions). However, you can't just simply put one cation and one anion together to form a compound. Each compound needs to been neutral, or have an overall charge of 0. When cations and anions do not have charges that perfectly cancel, you need to modify the amount of each ion in the compound.
1.) Fe(CN)₂
-----> Fe²⁺ and CN⁻
-----> +2 + (-1) + (-1) = 0
2.) FeCO₃
-----> Fe²⁺ and CO₃²⁻
-----> +2 + (-2) = 0
3.) Pb(CN)₄
-----> Pb⁴⁺ and CN⁻
-----> +4 + (-1) + (-1) + (-1) + (-1) = 0
4.) Pb(CO₃)₂
-----> Pb⁴⁺ and CO₃²⁻
-----> +4 +(-2) + (-2) = 0
Answer:
The standard cell potential of the reaction is 0.78 Volts.
Explanation:

Reduction at cathode :
Reduction potential of
to Cu=
Oxidation at anode:

Reduction potential of
to Fe=
To calculate the
of the reaction, we use the equation:

Putting values in above equation, we get:

The standard cell potential of the reaction is 0.78 Volts.
The mass number of an isotope can be expressed <span>by simply writing the name of the element or symbol followed by a hyphen and the mass number.
Example:
Carbon-13, Carbon-14
Oxygen-17
Uranium-235
</span>
<span>This
really depends on how closely related the species are. Species from vastly
unrelated taxonomic groups are likely to have organs that differ substantially.
Think for example of the compound eye of a spider and the eye of a human, or
the bones of a fish compared to the cartilage of a shark. These are examples of species that are not closely related at all. Then think of a chimpanzee and a human. The organs of both species are very similar in form and function as they are closely related. </span>